Question: Divide the following complex numbers. $ \dfrac{1+21i}{-3+2i}$
Answer: We can divide complex numbers by multiplying both numerator and denominator by the denominator's complex conjugate , which is ${-3-2i}$ $ \dfrac{1+21i}{-3+2i} = \dfrac{1+21i}{-3+2i} \cdot \dfrac{{-3-2i}}{{-3-2i}} $ We can simplify the denominator using the fact $(a + b) \cdot (a - b) = a^2 - b^2$ $ \dfrac{(1+21i) \cdot (-3-2i)} {(-3+2i) \cdot (-3-2i)} = \dfrac{(1+21i) \cdot (-3-2i)} {(-3)^2 - (2i)^2} $ Evaluate the squares in the denominator and subtract them. $ \dfrac{(1+21i) \cdot (-3-2i)} {(-3)^2 - (2i)^2} = $ $ \dfrac{(1+21i) \cdot (-3-2i)} {9 + 4} = $ $ \dfrac{(1+21i) \cdot (-3-2i)} {13} $ Note that the denominator now doesn't contain any imaginary unit multiples, so it is a real number, simplifying the problem to complex number multiplication. Now, we can multiply out the two factors in the numerator. $ \dfrac{({1+21i}) \cdot ({-3-2i})} {13} = $ $ \dfrac{{1} \cdot {(-3)} + {21} \cdot {(-3) i} + {1} \cdot {-2 i} + {21} \cdot {-2 i^2}} {13} $ Evaluate each product of two numbers. $ \dfrac{-3 - 63i - 2i - 42 i^2} {13} $ Finally, simplify the fraction. $ \dfrac{-3 - 63i - 2i + 42} {13} = \dfrac{39 - 65i} {13} = 3-5i $